If
the region is bounded by the x-axis and the graph y = f(x) and the
solid is obtained by rotating the region about the x-axis, then we
need to find the integral π∫_{a}^{b}
f^{2}(x)
dx, where either the interval a ≤ x ≤ b is given or its endpoints
are found as x-intercepts of the graph y = f(x).

In
case when the region is bounded by the vertical lines x = a, x = b,
and the graphs y = f_{1}(x)
and y=f_{2}(x),
where f_{1}(x)
≤ f_{2}(x)
for a ≤ x ≤ b, the volume of the solid obtained by revolution of
the region about the x-axis equals π∫_{a}^{b}
[f_{2}^{2}(x)
‒ f_{1}^{2}(x)]dx.

PS. If you have specific questions about this volume, please let me know.